$\int^{1}_{0}\dfrac{4x}{\sqrt{1-x^4}}\,dx\, = $
Answer: Strategy Let's first find the indefinite integral $\int\dfrac{4x}{\sqrt{1-x^4}}\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\dfrac{4x}{\sqrt{1-x^4}}\,dx\, $, we can use U-substitution. If we let $ {u=x^2}$, then ${du=2x \, dx}$ and ${ 4x\,dx=2\, du}$. So we have: $\begin{aligned}\int\dfrac{4x}{\sqrt{1-x^4}}\,dx\, &=\int\dfrac{1}{\sqrt{1-({x^2})^2}}\cdot {4x\,dx\,} \\\\\\\\ &=\int\dfrac{1}{\sqrt{1- u^2}}\,\cdot {2\, du}\,\\\\\\\\ &=2\int \dfrac{1}{\sqrt{1-u^2}}\,du\\\\\\\\ &=2 \sin^{-1}(u)+C\\\\\\\\ &=2 \sin^{-1}(x^2)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{1}_{0}\dfrac{4x}{\sqrt{1-x^4}}\,dx\,&= 2 \sin^{-1}(x^2)\Bigg|^1_0\\\\\\\\ &=2\left(\sin^{-1}(1)-\sin^{-1}(0)\right)\\\\\\\\ &=2\left(\dfrac{\pi}{2}-0\right)\\\\\\\\ &=\pi\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{1}_{0}\dfrac{4x}{\sqrt{1-x^4}}\,dx\, = \pi$